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satellite73Best ResponseYou've already chosen the best response.1
method one solve \[x^3=8\] \[x^38=0\] \[(x2)(x^2+2x+4)=0\] quadratic formula will give the solutions to \[x^2+2x+4=0\]
 one year ago

scottmanBest ResponseYou've already chosen the best response.0
I see and so 2 is one root and the other 2 complex numbers will be the other 2 roots?
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
method two solve \(x^3=1\) and multiply by two since one solution is 1, divide the unit circle in two three equal parts one is at 1, the other two are \(\frac{1}{2}+\frac{\sqrt{3}}{2}i\) and its conjugate \(\frac{1}{2}\frac{\sqrt{3}}{2}i\) multiply these numbers by 2
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
to answer your question, yes, the solution to the quadratic will give you the two complex roots
 one year ago

scottmanBest ResponseYou've already chosen the best response.0
ok thanks a lot, you helped me greatly
 one year ago
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